Question

Find the inverse of each of the following matrices by using elementary row transformations:

$\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$

$$\begin{gathered} A=IA \\ \Rightarrow \left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \mathrm{Applying}{R}_1\leftrightarrow R_3 \\ \Rightarrow \left[\begin{array}{ccc}1& 1& -2\\ 3& 2& -4\\ 2& -3& 5\end{array}\right]=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]A \\ \mathrm{Applying}{R}_2\to R_2-3R_1\mathrm{and}{R}_3\to R_3-2R_1 \\ \Rightarrow \left[\begin{array}{ccc}1& 1& -2\\ 0& -1& 2\\ 0& -5& 9\end{array}\right]=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& -3\\ 1& 0& -2\end{array}\right]A \\ \mathrm{Applying}{R}_2\to \left(-1\right)R_2 \\ \Rightarrow \left[\begin{array}{ccc}1& 1& -2\\ 0& 1& -2\\ 0& -5& 9\end{array}\right]=\left[\begin{array}{ccc}0& 0& 1\\ 0& -1& 3\\ 1& 0& -2\end{array}\right]A \\ \mathrm{Applying}{R}_3\to R_3+5R_2\mathrm{and}{R}_1\to R_1-R_2 \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& -2\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ 0& -1& 3\\ 1& -5& 13\end{array}\right]A \\ \mathrm{Applying}{R}_3\to \left(-1\right)R_3 \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& -2\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ 0& -1& 3\\ -1& 5& -13\end{array}\right]A \\ \mathrm{Applying}{R}_2\to R_2+2R_3 \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]A \\ \mathrm{Hence},A^{-1}=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right] \end{gathered}$$